Natural Numbers form Inductive Set

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Theorem

Let $\N$ denote the natural numbers as subset of the real numbers $\R$.


Then $\N$ is an inductive set.


Proof 1

By definition of the natural numbers:

$\N = \displaystyle \bigcap \mathcal I$

where $\mathcal I$ is the collection of all inductive sets.


Suppose that $n \in \N$.

Then by definition of intersection:

$\forall I \in \mathcal I: n \in I$

Since all these $I$ are inductive:

$\forall I \in \mathcal I: n + 1 \in I$

Again by definition of intersection:

$n + 1 \in \N$

$\blacksquare$


Proof 2

By the given definition of the natural numbers:

$\N = \displaystyle \bigcap \mathcal I$

where $\mathcal I$ is the collection of all inductive sets.


The result is a direct application of Intersection of Inductive Sets is Inductive Set.

$\blacksquare$


Sources